Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
You are given a 0-indexed integer array nums and an integer k. You have a starting score of 0.
In one operation:
i such that 0 <= i < nums.length,nums[i], andnums[i] with ceil(nums[i] / 3).Return the maximum possible score you can attain after applying exactly k operations.
The ceiling function ceil(val) is the least integer greater than or equal to val.
Example 1:
Input: nums = [10,10,10,10,10], k = 5 Output: 50 Explanation: Apply the operation to each array element exactly once. The final score is 10 + 10 + 10 + 10 + 10 = 50.
Example 2:
Input: nums = [1,10,3,3,3], k = 3 Output: 17 Explanation: You can do the following operations: Operation 1: Select i = 1, so nums becomes [1,4,3,3,3]. Your score increases by 10. Operation 2: Select i = 1, so nums becomes [1,2,3,3,3]. Your score increases by 4. Operation 3: Select i = 2, so nums becomes [1,2,1,3,3]. Your score increases by 3. The final score is 10 + 4 + 3 = 17.
Constraints:
1 <= nums.length, k <= 1051 <= nums[i] <= 109Problem summary: You are given a 0-indexed integer array nums and an integer k. You have a starting score of 0. In one operation: choose an index i such that 0 <= i < nums.length, increase your score by nums[i], and replace nums[i] with ceil(nums[i] / 3). Return the maximum possible score you can attain after applying exactly k operations. The ceiling function ceil(val) is the least integer greater than or equal to val.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Greedy
[10,10,10,10,10] 5
[1,10,3,3,3] 3
sliding-window-maximum)remove-stones-to-minimize-the-total)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2530: Maximal Score After Applying K Operations
class Solution {
public long maxKelements(int[] nums, int k) {
PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);
for (int v : nums) {
pq.offer(v);
}
long ans = 0;
while (k-- > 0) {
int v = pq.poll();
ans += v;
pq.offer((v + 2) / 3);
}
return ans;
}
}
// Accepted solution for LeetCode #2530: Maximal Score After Applying K Operations
func maxKelements(nums []int, k int) (ans int64) {
h := hp{nums}
heap.Init(&h)
for ; k > 0; k-- {
ans += int64(h.IntSlice[0])
h.IntSlice[0] = (h.IntSlice[0] + 2) / 3
heap.Fix(&h, 0)
}
return
}
type hp struct{ sort.IntSlice }
func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
func (hp) Push(any) {}
func (hp) Pop() (_ any) { return }
# Accepted solution for LeetCode #2530: Maximal Score After Applying K Operations
class Solution:
def maxKelements(self, nums: List[int], k: int) -> int:
h = [-v for v in nums]
heapify(h)
ans = 0
for _ in range(k):
v = -heappop(h)
ans += v
heappush(h, -(ceil(v / 3)))
return ans
// Accepted solution for LeetCode #2530: Maximal Score After Applying K Operations
use std::collections::BinaryHeap;
impl Solution {
pub fn max_kelements(nums: Vec<i32>, k: i32) -> i64 {
let mut pq = BinaryHeap::from(nums);
let mut ans = 0;
let mut k = k;
while k > 0 {
if let Some(v) = pq.pop() {
ans += v as i64;
pq.push((v + 2) / 3);
k -= 1;
}
}
ans
}
}
// Accepted solution for LeetCode #2530: Maximal Score After Applying K Operations
function maxKelements(nums: number[], k: number): number {
const pq = new MaxPriorityQueue<number>();
nums.forEach(num => pq.enqueue(num));
let ans = 0;
while (k > 0) {
const v = pq.dequeue();
ans += v;
pq.enqueue(Math.floor((v + 2) / 3));
k--;
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.
Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.